Ans. Since 1 the root of the quadratic equation kx2 - 7x + 5 = 0
∴ it will satisfy the given quadratic equation
∴ Substituting x = 1 in given quadratic equation
∴ kx2 - 7x + 5 = 0
∴ k(1)2 - 7(1) + 5 = 0
∴ k - 7 + 5 = 0
∴ k - 2 = 0
∴ k = 2.
3. Attempt any three of the following sub questions:
1. Area under different crops in a certain village is given below. Represent it by a pie diagram. [VIDEO]
Crop
Area in Hectares
Jowar
40
Wheat
60
Sugarcane
50
Vegetables
30
Ans.
Crop
Area in Hectares
Measure of Central Angle (Θ)
Jowar
40
(40 ÷ 180) × 360 = 800
Wheat
60
(60 ÷ 180) × 360 = 1200
Sugarcane
50
(50 ÷ 180) × 360 = 1000
Vegetables
30
(30 ÷ 180) × 360 = 600
Total
180
3600
2. If two coins are tossed, then find the probability of the event that at the most one tail turns up. [VIDEO]
Ans. Since two coins are tossed,
∴ The Sample Space is,
S = { HH, HT, TH, TT}
∴ n(S) = 4
Let A = Event of getting at the most one tail turns up in a toss.
∴ A = {HH, HT, TH}
∴ n(A) = 3
∴ P(A) = n(A) ÷ n(S)
∴ P(A) = 3 ÷ 4
∴ P(A) = ¾
3. Solve the following simultaneous equations using graphical method. [VIDEO]
x + y = 7;
x - y = 5.
Ans. x + y = 7 [Given]
Let us substitute the value of x as 0, 1 and 2 in the given simultaneous equation.
We will get the value of y as 7, 6 and 5 respectively.
x
0
1
2
y
7
6
5
( x , y )
( 0, 7 )
( 1, 6 )
( 2, 5 )
Now, we take the second equation x - y = 5,
Here we substitute the value of y as 0, 1 and 2 then we will get the value of x as 5, 6 and 7 respectively.
x
5
6
7
y
0
1
2
( x , y )
( 5, 0 )
(6, 1)
(7, 2)
4. There is an auditorium with 35 rows of seats. There are 20 seats in the first row, 22 seats in the second row, 24 seats in the third row and so on. Find the number of seats in the twenty second row. [VIDEO]
Ans. Since the no. of seats in each row of the auditorium are 20, 22, 24, ......
The no. of seats in each row form an A.P.
No. of seats in first row (a) = 20
Difference in no. of seats in two successive rows is (d) = 2
No. of seats in 22nd row = t22 = ?
tn = a + (n + 1) d
∴ t22 = a + (22 – 1) d
∴ t22 = 20 + 21 (2)
∴ t22 = 20 + 42
∴ t22 = 62
∴ There are 62 seats in 22th row.
5. Solve the following quadratic equation by completing square method.
Q is the event that the number so formed is greater than 50
Q = { 51, 52, 53, 54 }
∴ n (Q) = 4
R is the event that number so formed is divisible by 3
R = { 12, 15, 21, 24, 30, 42, 45, 51, 54 }
∴ n (R) = 9
2. The following table shows ages of 300 patients getting medical treatment in a hospital on a particular day.
Age ( in years)
No. of Patients.
10 - 20
60
20 - 30
42
30 - 40
55
40 - 50
70
50 - 60
53
60 - 70
20
Find the median age of the patients.
Ans.
3. If α + β = 5 and α3 + β3 = 35, find the quadratic equation whose roots are α and β.
Ans. Let α and β be the roots of a quadratic equation.
∴ α + β = 5 [Given] ...... eq. (1)
and α3 + β3 = 35 ............ eq. (2)
We know that,
α3 + β3 = (α + β )3 – 3αβ (α + β )
∴ 35 = 53 – 3αβ (5) [ From eq. (1) & (2)]
∴ 35 – 125 = - 15αβ
∴ -90 = - 15αβ
∴ α β = - 90/-15
∴ α β = 6
We know that, Quadratic equation is given by,
x2 – (Sum of the roots)x + Product of the roots = 0
∴ x2 –( α + β)x + αβ = 0
∴ x2 – 5x + 6 = 0 is the required quadratic equation.
5. Attempt any two of the following subquestions: [10]
1. Babubhai borrows Rs. 4000 and agrees to repay with a total interest of Rs, 500 in 10 instalments, each instalment being less than the preceding instalment by Rs. 10. What should be the first and the last instalments?
Ans. Total money repaid by Babubhai in 10 instalments = (S10)= Principle + Interest
∴ S10 = 4000 + 500
∴ S10 = Rs. 4500
No. of instalments (n) = 10
∵ He repays each instalment being less than the preceding instalment by Rs. 10.
∴ Difference between two consecutive instalments (d) = – 10
First instalment = (a) = ?
Last instalment (t10) = ?
We know that,
Sn = n/2[2a + (n – 1) d]
∴ S10 = 10/2 [2a + (10 – 1) d]
∴ 4500 = 5 [2a + 9 (– 10)]
∴ 4500/5 = 2a – 90
∴ 900 = 2a – 90
∴ 900 + 90 = 2a
∴ 990 = 2a
∴ a = 990/2
∴ a = 495 ............. eq. (i)
Now, tn = a + (n – 1) d
∴ t10 = a + (10 – 1) d
∴ t10 = 495 + 9 (– 10) [From eq. (i)]
∴ t10 = 495 – 90
∴ t10 = 405
∴ First instalment is Rs. 495 and last instalment is Rs.405
2. On the first day of the sale of tickets of a drama, in all 35 tickets were sold. If the rates of the tickets were Rs. 20 and Rs. 40. per ticket and the total collection was Rs. 900, find the number of tickets sold of each rate.
Ans. Let the number of tickets sold at the rate of Rs. 20 be x and at the rate of Rs. 40 be y.
According to first condition,
No. of tickets sold @ Rs. 20 + No. of tickets sold @ Rs. 40 = Total No. of Tickets sold (i.e.) 35 tickets
∴ x + y = 35 _____________ eq. (1)
According to second condition,
Amount received by selling Rs. 20 tickets + Amount Received by selling Rs. 40 Tickets = Total Collection received (i.e.) Rs. 900
∴ 20x +40y = 900
∴ 20(x + 2y) = 900
∴ x + 2y = 900/20
∴ x + 2y = 45 ________________ eq. no. (2)
Subtracting Equation (1) From Equation (2)
x + 2y = 45
x + y = 35
(-) (-) (-)
__________
y = 10
_________
Substituting y = 15 in equation (1)
∴ x + y = 35
∴ x + 10 = 35
∴ x = 35 - 10
∴ x = 25
∴ 25 no. of Rs. 20 Tickets and 10 no. of Rs. 40 Tickets were sold on the first day of the sale of tickets of a drama.
3. Given below is the frequency distribution of driving speeds (in km/hour) of the vehicles of 400 college students:
Speed ( in km/hr)
No. of Students
20 - 30
6
30 - 40
80
40 - 50
156
50 - 60
98
60 - 70
60
Draw Histogram and hence the frequency polygon for the above data.